91Ó°ÊÓ

An engineer whose mass is $m=70\text{kg}$holds onto the outer rim of a rotating space station whose radius is $R=14\text{m}$and which takes $t=30\text{s}$to make one complete rotation.

A force is a push or pull on an object because of the interaction of the thing with another object. Every time two things interact, a force is exerted on each of them. The acted force may be of attraction or repulsion. The two items no longer feel the force after the interaction ends.

The net force on an object is equal to the rate of change of momentum and can be written as the sum of the two components.

The parallel rate of change of momentum${\frac{d\stackrel{→}{p}}{dt}}_{\left|\right|}$and the perpendicular rate of change of momentum${\frac{d\stackrel{→}{p}}{dt}}_{âŠ\yen }$are two elements that we are concerned with.

So, the net force$F_{net}$on the object is given by,

$$\begin{gathered} F_{net}=\frac{d\stackrel{→}{p}}{dt} \\ ={\frac{d\stackrel{→}{p}}{dt}}_{\left|\right|}+{\frac{d\stackrel{→}{p}}{dt}}_{âŠ\yen } \end{gathered}$$

The engineer’s speed is affected by the parallel rate of change of momentum and given the speed is constant, therefore, the parallel rate is zero, and it equals the size of momentum rate change.

${\frac{d\stackrel{→}{p}}{dt}}_{\left|\right|}=0$

The centrifugal force$F_c$equals the rate change, which is the change in the direction owing to a perpendicular rate of change.

At speeds significantly slower than the speed of light, the magnitude of the perpendicular rate change is given by,

$$\begin{gathered} F_c={\frac{d\stackrel{→}{p}}{dt}}_{âŠ\yen } \\ =\frac{m{v}^2}{R} \end{gathered}$$

The change in distance over time is the speed.

So, it is given by,

$v=\frac{d}{t}$

The engineer goes above the diameter of a circle as he travels through one cycle.

The circumference is given by,

$2\mathrm{Ï€}R$

Where, R is the radius of the circle.

So, the distance where the engineer travel is:

$d=2\mathrm{Ï€}R$

Therefore,

$v=\frac{2\mathrm{Ï€}R}{t}$

Now, put the values for R and t to get the speed of the engineer

$$\begin{gathered} v=\frac{2\mathrm{Ï€}R}{t} \\ =\frac{2\mathrm{Ï€}\left(14\text{m}\right)}{30\text{s}} \\ =2.94\text{m}/\text{s} \end{gathered}$$

Now, put the values for $m,v,andRtoget{F}_c$

$$\begin{gathered} F_c=\frac{m{v}^2}{R} \\ =\frac{\left(70\text{kg}\right){(2.94\text{m}/\text{s})}^2}{14\text{m}} \\ =43.21\text{N} \end{gathered}$$

The magnitude of the force the engineer has to exert to hold on is$43.21N$_.

At speeds slower than the speed of light, the net force applied on the object equals the rate of change of momentum and the magnitude of the perpendicular rate change, which is the centrifugal force$F_c$.

$$\begin{gathered} F_{net}=F_c \\ ={\frac{d\stackrel{→}{p}}{dt}}_{âŠ\yen } \\ =43.21\text{N} \end{gathered}$$

Thus, the magnitude of the net force acting on the engineer is $43.21N$.